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A-Level · Further Statistics (Discrete Distributions Depth: Poisson, Geometric and Negative Binomial via Probability Generating Functions)

FP.FS5 Further Statistics: Discrete Distributions Depth

EDEXCEL 9FM0 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Key facts: discrete distributions and probability generating functions

Original content written for Revision Library.

For X ~ Po(lambda), P(X = x) = e^(-lambda) * lambda^x / x!, for x = 0, 1, 2, ..., with E(X) = lambda and Var(X) = lambda. If X ~ Po(lambda1) and Y ~ Po(lambda2) are independent, then X + Y ~ Po(lambda1 + lambda2).

For X ~ Geo(p), the number of independent trials up to and including the first success (write q = 1 - p), P(X = x) = q^(x-1) * p, for x = 1, 2, 3, ..., with E(X) = 1/p and Var(X) = q/p^2. The geometric distribution has the memoryless property: P(X > m+n | X > m) = P(X > n) for positive integers m and n.

For X ~ NB(r, p), the number of independent trials up to and including the rth success, P(X = x) = C(x-1, r-1) * p^r * q^(x-r), for x = r, r+1, r+2, ..., with E(X) = r/p and Var(X) = rq/p^2. X can be written as the sum of r independent Geo(p) random variables.

The probability generating function of a discrete random variable X taking non-negative integer values is G_X(t) = E(t^X) = sum of P(X = x) * t^x. It satisfies G_X(1) = 1, E(X) = G_X'(1), and Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2. If X and Y are independent random variables, then G_(X+Y)(t) = G_X(t) * G_Y(t).

A call centre in Norwich receives incoming telephone calls singly and at random, independently of each other, at a constant average rate of 3.6 calls per 10-minute period. Let X be the number of calls received during a randomly chosen 10-minute period, so that X ~ Po(3.6).

(a)Find P(X = 4), giving your answer to 3 significant figures.(2)

(b)Find P(X ≥ 2), giving your answer to 3 significant figures.(2)

(Total for Question 1 is 4 marks)

Meera is practising penalty kicks. Each attempt is successful, independently of all other attempts, with probability 0.35. Let X be the number of attempts up to and including her first successful kick, so that X ~ Geo(0.35).

(a)Find P(X = 5), giving your answer to 3 significant figures.(2)

(b)Find P(X > 3), giving your answer to 3 significant figures.(2)

(c)State the value of E(X), giving your answer as an exact fraction, and interpret this value in the context of this scenario.(1)

(Total for Question 2 is 5 marks)

In a factory in Swansea, each electronic component produced is defective, independently of all other components, with probability 0.08. Components are inspected one at a time until the 3rd defective component is found. Let X be the total number of components inspected up to and including the 3rd defective one found, so that X follows a negative binomial distribution, X ~ NB(3, 0.08), where P(X = x) = C(x-1, 2) * (0.08)³ * (0.92)^x-3, for x = 3, 4, 5, ... .

(a)Write down the value of E(X).(1)

(b)Find the probability that the 3rd defective component found is the 10th component inspected, giving your answer to 3 significant figures.(3)

(c)Find Var(X).(1)

(Total for Question 3 is 5 marks)

The discrete random variable X has the probability distribution shown in the table below.
X: 0, 1, 2
P(X = x): 0.2, 0.3, 0.5

(a)Write down the probability generating function G_X(t) of X.(2)

(b)Verify that G_X(1) = 1, and explain why this must always be true for the probability generating function of any discrete random variable.(2)

(c)Use G_X(t) to find E(X).(2)

(Total for Question 4 is 6 marks)

The discrete random variable X follows a Poisson distribution with parameter λ, so that X ~ Po(λ), where P(X = x) = e^-λ * λ^x / x!, for x = 0, 1, 2, ... .

(a)Show that the probability generating function of X is given by G_X(t) = e^λ(t-1).(4)

(b)Using your result from part (a), show that E(X) = λ.(3)

(c)Using your result from part (a), show further that Var(X) = λ.(3)

(Total for Question 5 is 10 marks)

The discrete random variable X follows a geometric distribution with parameter p, so that X ~ Geo(p), where P(X = x) = (1-p)^x-1 * p, for x = 1, 2, 3, ... . Write q = 1 - p.

(a)Show that the probability generating function of X is G_X(t) = pt / (1 - qt), for |t| < 1/q.(4)

(b)Hence show that E(X) = 1/p.(3)

(c)Hence show further that Var(X) = q/p^2.(4)

(Total for Question 6 is 11 marks)

Let X ~ Po(lambda1) and Y ~ Po(lambda2) be independent random variables, and let W = X + Y.

(a)State the probability generating functions of X and of Y.(1)

(b)Show that G_W(t) = G_X(t) * G_Y(t), and hence show that W ~ Po(lambda1 + lambda2).(4)

(c)The number of £10 notes given as change by the till of a small shop in Dundee during the morning session follows a Poisson distribution with mean 2.4, independently of the number given as change during the afternoon session, which follows a Poisson distribution with mean 3.1. Find the probability that exactly 5 £10 notes are given as change over the whole day, giving your answer to 3 significant figures.(3)

(Total for Question 7 is 8 marks)

A sequence of independent trials is carried out, in each of which the probability of success is p (write q = 1 - p). Let X ~ NB(r, p) be the number of trials up to and including the rth success. X can be written as X = X1 + X2 + ... + Xr, where X1 is the number of trials up to and including the 1st success, and, for i = 2, 3, ..., r, Xi is the number of further trials after the (i-1)th success up to and including the ith success. X1, X2, ..., Xr are independent, and each has the Geo(p) distribution.

(a)State the probability generating function of a single Xi.(1)

(b)Hence show that the probability generating function of X is G_X(t) = [pt / (1 - qt)]^r.(3)

(c)Using G_X(t), show that E(X) = r/p.(4)

(d)Show further that Var(X) = rq/p^2.(4)

(Total for Question 8 is 12 marks)

Let X ~ Geo(p) be the number of trials up to and including the first success, and write q = 1 - p.

(a)Show that, for positive integers m and n, P(X > m+n | X > m) = P(X > n).(4)

(b)A component is tested repeatedly until it fails; the probability that it fails on any given test is 0.04, independently of the outcome of every previous test. Given that the component has already survived (not failed) 10 tests, find the probability that it survives at least a further 15 tests before it fails, giving your answer to 3 significant figures.(2)

(Total for Question 9 is 6 marks)

In parts (a) and (b) below, X ~ Geo(p) models the number of independent trials, each with probability of success p, up to and including the first success; the two parts describe separate, independent scenarios.

(a)Given that E(X) = 12.5, find the value of p.(2)

(b)Given instead that Var(X) = 20, form an equation in p and show that it can be written as 20p² + p - 1 = 0. Hence find the value of p, given that 0 < p < 1.(5)

(Total for Question 10 is 7 marks)

The discrete random variable Y has the probability distribution shown in the table below.
Y: 1, 2, 3, 4
P(Y = y): 0.1, 0.4, 0.3, 0.2

(a)Find the probability generating function G_Y(t).(2)

(b)Use G_Y(t) to find E(Y) and Var(Y).(5)

(c)Verify your value of E(Y) by direct calculation from the probability distribution.(2)

(Total for Question 11 is 9 marks)

Small pieces of grit land on a stretch of pavement in Aberdeen singly and at random, independently of each other, at a constant average rate of 0.6 pieces of grit per metre length.

(a)Let X be the number of pieces of grit found on a randomly chosen 5 metre length of pavement. State the distribution of X, including the value of any parameter(s).(2)

(b)Find P(X = 3), giving your answer to 3 significant figures.(2)

(c)Find P(X ≥ 1), giving your answer to 3 significant figures.(2)

(d)Two non-overlapping stretches of pavement, of lengths 5 metres and 8 metres, are considered. Using a suitable property of the Poisson distribution, find the probability that the total number of pieces of grit found on the combined 13 metre stretch is at most 5, giving your answer to 3 significant figures.(5)

(Total for Question 12 is 11 marks)

Mark scheme · FP.FS5 Further Statistics: Discrete Distributions Depth

Question 1

(a) M1 correct use of P(X=4) = e^-3.6 * 3.6⁴ / 4! (or calculator Poisson pdf), oe
(a) A1 awrt 0.191
(a) Answer: P(X = 4) = awrt 0.191
(b) M1 uses P(X≥2) = 1 - P(X=0) - P(X=1), oe
(b) A1 awrt 0.874
(b) Answer: P(X ≥ 2) = awrt 0.874

Question 2

(a) M1 correct use of P(X=5) = q⁴ * p with q = 0.65, p = 0.35, oe
(a) A1 awrt 0.0625
(a) Answer: P(X = 5) = awrt 0.0625
(b) M1 recognises P(X>n) = qⁿ, i.e. the probability the first 3 attempts all fail
(b) A1 0.274625, awrt 0.275
(b) Answer: P(X > 3) = 0.274625, awrt 0.275
(c) B1 E(X) = 1/0.35 = 20/7, awrt 2.86, with a valid interpretation in context, oe
(c) Answer: E(X) = 20/7 = awrt 2.86; on average, Meera needs about 2.86 attempts up to and including her first success.

Question 3

(a) B1 E(X) = r/p = 3/0.08 = 37.5, cao
(a) Answer: E(X) = 37.5
(b) M1 identifies x = 10, r = 3 and uses P(X=x) = C(x-1,r-1) * p^r * q^x-r, oe
(b) A1 correct unsimplified expression C(9,2) * (0.08)³ * (0.92)⁷, oe
(b) A1 awrt 0.0103
(b) Answer: P(X = 10) = awrt 0.0103
(c) B1 Var(X) = r*q/p² = 3*0.92/0.0064 = 431.25, cao
(c) Answer: Var(X) = 431.25

Question 4

(a) M1 G_X(t) = sum of P(X=x) * t^x, using the probabilities given
(a) A1 G_X(t) = 0.2 + 0.3t + 0.5t², oe
(a) Answer: G_X(t) = 0.2 + 0.3t + 0.5t²
(b) B1 G_X(1) = 0.2 + 0.3 + 0.5 = 1
(b) B1 valid explanation, e.g. G_X(1) = sum of all the probabilities in the distribution, which must total 1 for any valid probability distribution, oe
(b) Answer: G_X(1) = 1, since the probabilities in any distribution must sum to 1.
(c) M1 differentiates: G_X'(t) = 0.3 + t, oe
(c) A1 E(X) = G_X'(1) = 1.3, cao
(c) Answer: E(X) = 1.3

Question 5

(a) M1 writes G_X(t) = sum from x=0 to infinity of t^x * e^-λ * λ^x / x!, oe
(a) M1 factors out e^-λ and combines terms: G_X(t) = e^-λ * sum from x=0 to infinity of (λ*t)^x / x!
(a) M1 recognises sum from x=0 to infinity of (λ*t)^x/x! = e^λ*t (the exponential series)
(a) A1 G_X(t) = e^-λ * e^λ*t = e^λ(t-1), cso
(a) Answer: G_X(t) = e^λ(t-1)
(b) M1 differentiates: G_X'(t) = λ * e^λ(t-1)
(b) dM1 substitutes t = 1 into G_X'(t)
(b) A1 E(X) = G_X'(1) = λ * e⁰ = λ, cso
(b) Answer: E(X) = λ
(c) M1 differentiates again: G_X''(t) = λ² * e^λ(t-1), so G_X''(1) = λ²
(c) M1 uses Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]² = λ² + λ - λ², oe
(c) A1 Var(X) = λ, cso
(c) Answer: Var(X) = λ

Question 6

(a) M1 writes G_X(t) = sum from x=1 to infinity of t^x * q^x-1 * p, oe
(a) M1 takes out a factor of pt: G_X(t) = pt * sum from k=0 to infinity of (qt)^k, oe
(a) M1 recognises sum from k=0 to infinity of (qt)^k = 1/(1-qt) as an infinite geometric series, valid for |qt| < 1
(a) A1 G_X(t) = pt/(1-qt), cso
(a) Answer: G_X(t) = pt/(1-qt)
(b) M1 differentiates using the quotient rule: G_X'(t) = p*(1-qt)^-1 + p*q*t*(1-qt)^-2, oe
(b) dM1 substitutes t = 1, using 1 - q = p
(b) A1 E(X) = G_X'(1) = p/p + pq/p² = 1 + q/p = (p+q)/p = 1/p, cso
(b) Answer: E(X) = 1/p
(c) M1 differentiates G_X'(t) again to obtain G_X''(t) = 2pq*(1-qt)^-2 + 2pq²*t*(1-qt)^-3, oe
(c) dM1 substitutes t = 1 to get G_X''(1) = 2q/p + 2q²/p²
(c) dM1 uses Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]² and combines all terms over a common denominator p²
(c) A1 Var(X) = q/p², cso, using p+q=1 to simplify the numerator to q
(c) Answer: Var(X) = q/p²

Question 7

(a) B1 G_X(t) = e^lambda1(t-1) and G_Y(t) = e^lambda2(t-1), both stated (ft from Q5, or quoted)
(a) Answer: G_X(t) = e^lambda1(t-1), G_Y(t) = e^lambda2(t-1)
(b) M1 writes G_W(t) = E(t^X+Y) = E(t^X * t^Y), oe
(b) M1 uses independence of X and Y: E(t^X * t^Y) = E(t^X)*E(t^Y) = G_X(t)*G_Y(t)
(b) M1 combines: G_W(t) = e^lambda1(t-1) * e^lambda2(t-1) = e^{(lambda1+lambda2)(t-1)}
(b) A1 identifies this as the PGF of a Po(lambda1+lambda2) distribution; since a PGF uniquely determines a distribution, W ~ Po(lambda1+lambda2), cso
(b) Answer: W ~ Po(lambda1 + lambda2)
(c) M1 identifies W ~ Po(2.4+3.1) = Po(5.5), using the result of part (b), oe
(c) M1 correct use of P(W=5) = e^-5.5 * 5.5⁵ / 5!
(c) A1 awrt 0.171
(c) Answer: P(W = 5) = awrt 0.171

Question 8

(a) B1 G_(Xi)(t) = pt/(1-qt), ft from Q6(a) or quoted
(a) Answer: G_(Xi)(t) = pt/(1-qt)
(b) M1 states X is the sum of r independent, identically distributed random variables
(b) M1 uses the PGF-of-a-sum result (extended to r independent variables): the PGF of a sum of independent variables is the product of their individual PGFs
(b) A1 G_X(t) = [pt/(1-qt)]^r, cso
(b) Answer: G_X(t) = [pt/(1-qt)]^r
(c) M1 lets g(t) = pt/(1-qt), so G_X(t) = [g(t)]^r; differentiates using the chain rule: G_X'(t) = r*[g(t)]^r-1 * g'(t)
(c) M1 uses g(1) = 1 and g'(1) = 1/p (from Q6)
(c) dM1 substitutes t = 1: G_X'(1) = r * 1^r-1 * (1/p)
(c) A1 E(X) = G_X'(1) = r/p, cso
(c) Answer: E(X) = r/p
(d) M1 differentiates again using the product/chain rule: G_X''(t) = r(r-1)*[g(t)]^r-2*[g'(t)]² + r*[g(t)]^r-1*g''(t)
(d) M1 substitutes t=1 using g(1)=1, g'(1)=1/p, g''(1)=2q/p+2q²/p² (from Q6) to get G_X''(1) = r(r-1)/p² + 2rq/p + 2rq²/p²
(d) dM1 uses Var(X) = G_X''(1) + G_X'(1) - [G_X'(1)]² and combines all terms over a common denominator p²
(d) A1 Var(X) = rq/p², cso, using p+q=1 to simplify the numerator to rq
(d) Answer: Var(X) = rq/p²

Question 9

(a) B1 states or derives P(X > k) = q^k for a positive integer k, oe
(a) M1 uses P(X>m+n | X>m) = P(X>m+n and X>m)/P(X>m) = P(X>m+n)/P(X>m), since {X>m+n} is a subset of {X>m}
(a) M1 substitutes: q^m+n/q^m
(a) A1 = qⁿ = P(X>n), as required, cso
(a) Answer: P(X > m+n | X > m) = qⁿ = P(X > n) (shown)
(b) M1 uses the memoryless property from part (a): required probability = P(X>15) = q¹⁵, oe
(b) A1 awrt 0.542
(b) Answer: 0.542 (3 s.f.)

Question 10

(a) M1 uses E(X) = 1/p = 12.5, oe
(a) A1 p = 0.08, cao
(a) Answer: p = 0.08
(b) M1 uses Var(X) = (1-p)/p² = 20, oe
(b) M1 rearranges to 1 - p = 20p²
(b) A1 20p² + p - 1 = 0, cso
(b) M1 solves the quadratic, e.g. p = (-1 ± √1+80)/40 = (-1 ± 9)/40
(b) A1 p = 0.2 (rejecting p = -0.25 as 0 < p < 1), cao
(b) Answer: p = 0.2

Question 11

(a) M1 G_Y(t) = sum of P(Y=y)*t^y, using the table
(a) A1 G_Y(t) = 0.1t + 0.4t² + 0.3t³ + 0.2t⁴, oe
(a) Answer: G_Y(t) = 0.1t + 0.4t² + 0.3t³ + 0.2t⁴
(b) M1 differentiates: G_Y'(t) = 0.1 + 0.8t + 0.9t² + 0.8t³
(b) A1 E(Y) = G_Y'(1) = 2.6, cao
(b) M1 differentiates again: G_Y''(t) = 0.8 + 1.8t + 2.4t², so G_Y''(1) = 5.0
(b) M1 uses Var(Y) = G_Y''(1) + G_Y'(1) - [G_Y'(1)]², oe
(b) A1 Var(Y) = 0.84, cao
(b) Answer: E(Y) = 2.6, Var(Y) = 0.84
(c) M1 uses E(Y) = sum of y*P(Y=y) = 1(0.1)+2(0.4)+3(0.3)+4(0.2)
(c) A1 = 2.6, matching part (b), cao
(c) Answer: E(Y) = 0.1+0.8+0.9+0.8 = 2.6, matching part (b).

Question 12

(a) M1 uses parameter = rate * length = 0.6 * 5
(a) A1 X ~ Po(3), cao
(a) Answer: X ~ Po(3)
(b) M1 correct use of P(X=3) = e^-3 * 3³/3! with λ = 3
(b) A1 awrt 0.224
(b) Answer: P(X = 3) = awrt 0.224
(c) M1 uses P(X≥1) = 1 - P(X=0), oe
(c) A1 awrt 0.950
(c) Answer: P(X ≥ 1) = awrt 0.950
(d) M1 recognises the counts on the two non-overlapping stretches are independent Poisson variables, with parameters 0.6*5=3 and 0.6*8=4.8
(d) M1 uses the additive property (established in Q7): total W ~ Po(3+4.8) = Po(7.8), oe
(d) M1 sums the Poisson probabilities P(W=0) + P(W=1) + ... + P(W=5), or uses a calculator cumulative Poisson function
(d) A1 correct unsimplified sum, or calculator value, of awrt 0.2103
(d) A1 awrt 0.210
(d) Answer: P(W ≤ 5) = awrt 0.210

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